
Bounded modern-physics survey · Interactive lesson
Electron spin and the Stern-Gerlach experiment
Every quantum number so far fell out of a boundary condition. Spin does not: we read it off a measurement. Silver atoms split into exactly two beams, the size of the split weighs their moment at one Bohr magneton, and the g factor near 2 that follows rules out any picture of a little spinning ball.
UPIII-10.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.
Build the model
Two spots where classical physics predicts a smear, and a moment no orbit can supply.
The key idea
Electron spin is an intrinsic angular momentum with quantum number s = ½: its component along whatever axis you measure is only ever +ħ/2 or −ħ/2, and it carries a magnetic moment μ = −gₛ(e/2mₑ)S with gₛ ≈ 2.0023.
Example. A silver atom's lone 5s electron has l = 0, so its whole moment is spin: μz = ∓9.28 × 10⁻²⁴ J T⁻¹. A gradient of 800 T m⁻¹ pushes on it with 7.43 × 10⁻²¹ N, towards weaker field for mₛ = +½ and stronger field for mₛ = −½.
The complete physical model and assumptions
A magnetic moment in a field has energy U = −μ⋅B. A uniform field only twists the moment into precession; a field whose z-component changes with height also pushes, with Fz = μz ∂Bz/∂z, so an atom's deflection through such a magnet reads out μz directly. Classically μz = μ cos θ takes every value between −μ and +μ for moments arriving at random, and the beam should paint a continuous band. Silver atoms, whose single 5s electron has l = 0 and whose other 46 electrons cancel, paint two separated spots and nothing in between. Orbital angular momentum cannot do that: it has 2l + 1 projections, always an odd number, and l = 0 gives one undeflected beam. Two beams mean a new angular momentum with 2s + 1 = 2, so s = ½, Sz = ±ħ/2 and |S| = (√3/2)ħ. The size of the split fixes |μz| ≈ μB, twice what an orbiting charge with the same angular momentum carries, so gₛ ≈ 2. This spin is intrinsic, a two-valued degree of freedom that no function of position can carry, and the literal spinning sphere fails twice: at the classical electron radius its equator would move at about 170c, and any rotor whose charge follows its mass has g = 1. The model has an edge of its own: Dirac's relativistic equation gives g = 2 exactly, and the measured 2.00232 needs quantum electrodynamics.
1A uniform field twists; only a gradient pushes
A moment μ in a field B has energy U = −μ⋅B and feels a torque τ = μ × B. In a uniform field that torque makes μ precess about B at a fixed angle, and the net force −∇U is zero however strong B is. A force needs a field that changes across the atom's path: F = ∇(μ⋅B). A Stern-Gerlach magnet shapes one pole to a knife edge so that Bz grows steeply with z, giving Fz = μz ∂Bz/∂z. Maxwell's ∇⋅B = 0 forbids Bz from changing with z unless another component changes too, so a transverse force term appears as well. It averages away. The moment precesses about the strong Bz at γB, with γ = 1.76 × 10¹¹ rad s⁻¹ T⁻¹ for an electron spin, so in a 1 T field it turns about 2 × 10⁶ times during a 73 μs transit. The transverse projection averages to zero while μz stays steady, which is why the experiment measures one component and nothing else.
2Classical moments predict a smear
Atoms leave the oven with moments pointing every way. Uniform over directions means every element of solid angle, sin θ dθ dφ, is equally likely, so cos θ is uniformly distributed between −1 and +1. The probability density of μz = μ cos θ is then flat: P(μz) = 1/(2μ) for |μz| ≤ μ, which is normalised because ∫P dμz over that range is 1. Precession keeps θ fixed, so each atom holds its μz all the way through the magnet, and since the deflection is proportional to μz the screen should show one even band with no gap. The silver deposit Stern and Gerlach made in 1922 showed two separated lines with a clear gap between them. Quantum mechanics replaces the flat density with two spikes: in an unpolarised beam each atom lands in the upper or the lower spot with probability ½, and the pattern builds up one random landing at a time. No atom is ever found in the middle.
3Two beams count 2s + 1 = 2
Silver has 47 electrons. Forty-six fill closed shells whose angular momenta cancel, leaving one 5s electron with l = 0 and so no orbital moment. Orbital angular momentum could not give two beams anyway: its projections mₗħ run in whole steps from −lħ to +lħ, so an even count of beams needs a half-integer angular momentum, which no integer l supplies. Stern and Gerlach read their lines through the older Bohr–Sommerfeld picture. Spin was proposed by Uhlenbeck and Goudsmit in 1925, and in 1927 Phipps and Taylor sent ground-state hydrogen, whose l = 0 is beyond dispute, through a gradient and still saw two beams. So the electron carries an extra angular momentum with 2s + 1 = 2: s = ½, Sz = ±ħ/2 and |S| = √(s(s + 1)) ħ = 0.866ħ. The largest projection is shorter than the vector, so the spin lies on a cone at 54.7° to the axis. Correspondence check: a moment of large quantum number j gives 2j + 1 spots whose spacing is 1/(2j) of the full spread, and as j grows they merge into the classical band. Spin one-half is the opposite extreme.
4The size of the split weighs the moment
Inside a magnet of length L the atom feels a constant force for a time L/v, so it leaves with a sideways offset ½a(L/v)² and a sideways velocity aL/v, where a = μz (∂Bz/∂z)/M. A drift D to the screen adds aLD/v². Take silver, M = 1.791 × 10⁻²⁵ kg, at 550 m s⁻¹ through 800 T m⁻¹ with L = 4.0 cm and D = 6.0 cm: a = 4.15 × 10⁴ m s⁻², some 4200 times g, and each beam lands 0.439 mm off axis, 0.877 mm from the other. Run the formula backwards and a measured separation returns |μz|; for silver the answer is 9.28 × 10⁻²⁴ J T⁻¹, one Bohr magneton μB = eħ/2mₑ to within 0.12%. An orbital moment is μz = −μB mₗ, which is g = 1, so a projection of ħ/2 would carry only μB/2 and split the beams half as far. The spin carries twice the moment per unit of angular momentum: gₛ ≈ 2.
5Intrinsic means not made of motion in space
The integer mₗ of the previous lessons came from a boundary condition: a wavefunction must return to itself when φ goes to φ + 2π, so e(imφ) needs integer m. Every function of position obeys that condition, so no motion of anything through space can produce mₛ = ±½. Spin is a separate, two-valued degree of freedom: the electron's state carries a two-component spinor χ alongside its ψ(r), and the spin operators act on χ alone. The spinning-ball picture fails on its own terms as well. A sphere of the classical electron radius holding ħ/2 would need its equator to move at more than a hundred times c, and collisions show no electron structure at any scale yet probed. Any rigid rotor whose charge is spread like its mass has μ = (q/2m)L exactly, which is g = 1, half the measured value. The word spin survives; the rotation does not.
6Two is Dirac's number; 2.0023 is QED's
Nothing in the Schrödinger equation predicts spin; it is added by hand. Dirac's 1928 relativistic equation for the electron produces it unasked: the wavefunction needs four components, carrying two spin states for the electron and two for its antiparticle, and the magnetic moment comes out with g = 2 exactly. Its low-speed limit, the Pauli equation, carries g = 2 into atomic physics. The measured value is gₛ = 2.00231930, and the extra 0.1% is the electron's coupling to the quantised electromagnetic field, through virtual photons it emits and reabsorbs. Schwinger's first-order result, aₑ = (gₛ − 2)/2 = α/2π = 0.0011614, already lands only 0.15% above the measured 0.0011597, and higher powers of α/π close the gap, one of the closest agreements between theory and measurement in physics. A Stern-Gerlach deflection cannot see the 0.1%; the Zeeman and fine-structure splittings of the next lesson carry it simply as gₛ ≈ 2.0023.
Key equations
Read the units, assumptions and limits before substituting.
Force on a moment in a field gradient
Fz = μz ∂Bz/∂z
A uniform field, however strong, exerts no net force, only a torque. So the deflection measures the projection μz, never the full |μ|.
Fz in N; μz in J T⁻¹; ∂Bz/∂z in T m⁻¹. From U = −μ⋅B, keeping the one term that survives precession about BSpin projection
Sz = mₛ ħ, mₛ = +½ or −½
Two beams fix s = ½. The length |S| = √(s(s + 1)) ħ = 0.866ħ exceeds the 0.5ħ projection, so the spin sits at 54.7° to the axis, never along it.
Sz in J s; ħ = 1.055 × 10⁻³⁴ J s. 2s + 1 = 2 values, where orbital angular momentum gives 2l + 1, always oddSpin magnetic moment
μₛ = −gₛ (e/2mₑ) S
Along z, μz = −gₛ μB mₛ = ∓9.285 × 10⁻²⁴ J T⁻¹, where μB = eħ/2mₑ = 9.274 × 10⁻²⁴ J T⁻¹: a full Bohr magneton for half a unit of ħ.
μₛ in J T⁻¹; e = 1.602 × 10⁻¹⁹ C; mₑ = 9.109 × 10⁻³¹ kg; S in J s; gₛ = 2.00232, dimensionlessStern-Gerlach deflection at the screen
z = [μz (∂Bz/∂z) / (M v²)] (L²/2 + L D)
A constant force for a time L/v, then straight flight. The 1/v² means slow atoms are deflected most, so a thermal beam draws each spot out into a streak.
z, L, D in m, with L the magnet length and D the drift to the screen; M the atom's mass in kg; v its speed in m s⁻¹Equator speed of a spinning sphere
veq = 5ħ / (4 mₑ r)
At the classical electron radius, 2.82 × 10⁻¹⁵ m, veq = 5.1 × 10¹⁰ m s⁻¹, about 171c: no classical rotor of that size can hold the spin.
rigid uniform sphere of radius r in m, with (2/5) mₑ r veq = ħ/2; veq in m s⁻¹Anomalous moment from QED
gₛ = 2 + α/π + …
α/π = 0.0023228 gives gₛ = 2.0023228 against the measured 2.0023193. The first QED term overshoots the anomaly by 0.15%; higher orders close the gap.
α = e²/(4πε₀ħc) = 1/137.036, dimensionless; aₑ = (gₛ − 2)/2 is the anomaly; Dirac's equation alone gives gₛ = 2Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the gradient to zero and the spots merge: a uniform field only turns a moment. Then take v from 800 to 400 m s⁻¹ and the separation grows fourfold, the 1/v² of a fixed force acting for a time L/v; lengthening the magnet from 3 to 6 cm multiplies it by 2.4.
Live results
- FORCE |Fz|
- 7.43 × 10⁻²¹ N
- ACCELERATION |az|
- 41.5 km s⁻²
- TIME IN MAGNET L/v
- 72.7 μs
- SPOT SEPARATION
- 0.877 mm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Field gradient ∂Bz/∂z: 800 T m⁻¹; Atom speed v: 550 m s⁻¹; Magnet length L: 4.0 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
“Spin up” is not an arrow pointing up. The first-year picture draws mₛ = +½ as a vector of length ħ/2 lying along the field, fixed before the atom enters the magnet, which then merely sorts the arrows. Both halves fail. The length is |S| = √(s(s + 1)) ħ = 0.866ħ, longer than any projection, so the vector can never lie along z, and Sₓ and Sy have no definite values while Sz does. Nor does the magnet read out a direction the atom already had. Send the mₛ = +½ beam through a second magnet turned to x and it splits evenly; send either half through a third magnet turned back to z and it splits evenly again, although every atom entering the second magnet had been found with mₛ = +½. Measuring Sₓ erased Sz. A spin state is a two-valued quantum state, not a hidden arrow the apparatus uncovers.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A beam of potassium atoms (M = 39.10 u, one 4s electron outside closed shells) crosses a Stern-Gerlach magnet in which Bz increases towards +z at 450 T m⁻¹. How many spots form, what force and acceleration does each atom feel, and which way do the mₛ = +½ atoms go?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Closed shells carry no net angular momentum, and a 4s electron has l = 0, so the only moment is that electron's spin: 2s + 1 = 2 spots.
- Its projection is μz = −gₛ μB mₛ = ∓1.00116 × 9.274 × 10⁻²⁴ J T⁻¹ = ∓9.285 × 10⁻²⁴ J T⁻¹, the upper sign for mₛ = +½.
- Force: |Fz| = |μz| ∂Bz/∂z = 9.285 × 10⁻²⁴ J T⁻¹ × 450 T m⁻¹ = 4.18 × 10⁻²¹ N.
- Mass: M = 39.10 × 1.661 × 10⁻²⁷ kg = 6.49 × 10⁻²⁶ kg, so |a| = 4.18 × 10⁻²¹ N ÷ 6.49 × 10⁻²⁶ kg = 6.44 × 10⁴ m s⁻², about 6600 times g. Gravity does not matter here.
- Direction: for mₛ = +½, μz < 0, so Fz = μz ∂Bz/∂z < 0. Those atoms move towards −z, where the field is weaker; the mₛ = −½ atoms move towards +z.
Answer. Two spots. Each atom feels 4.18 × 10⁻²¹ N and accelerates at 6.44 × 10⁴ m s⁻²; mₛ = +½ atoms go towards −z, the weaker field, and mₛ = −½ atoms towards +z.
MediumExample 2
Silver atoms (M = 107.87 u) at 620 m s⁻¹ cross a 3.50 cm magnet with ∂Bz/∂z = 900 T m⁻¹ and condense on a plate at its exit. The two deposits are 0.150 ± 0.006 mm apart. Taking v, the gradient and L as exact, find |μz| in Bohr magnetons and the g factor, then test the result against a g = 1 moment with projection ħ/2.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Each beam leaves the magnet at z = ½a(L/v)² with a = |μz|(∂Bz/∂z)/M, and the two go opposite ways, so the separation is Δ = |μz|(∂Bz/∂z)L²/(Mv²). Invert: |μz| = ΔMv²/((∂Bz/∂z)L²).
- M = 107.87 × 1.6605 × 10⁻²⁷ kg = 1.7912 × 10⁻²⁵ kg, so Mv² = 1.7912 × 10⁻²⁵ kg × (620 m s⁻¹)² = 6.885 × 10⁻²⁰ J; and (∂Bz/∂z)L² = 900 T m⁻¹ × (0.0350 m)² = 1.1025 T m.
- |μz| = 1.50 × 10⁻⁴ m × 6.885 × 10⁻²⁰ J ÷ 1.1025 T m = 9.37 × 10⁻²⁴ J T⁻¹. Divided by μB = 9.274 × 10⁻²⁴ J T⁻¹ that is 1.010 μB, and the 4.0% uncertainty in Δ carries straight through: (1.01 ± 0.04) μB.
- With |mₛ| = ½, |μz| = gₛ μB/2, so gₛ = 2|μz|/μB = 2.02 ± 0.08, consistent with 2.0023, which predicts Δ = 0.149 mm.
- A g = 1 moment with projection ħ/2 has |μz| = μB/2 and would give Δ = 0.0742 mm. The measurement sits 0.076 mm, about 13 standard uncertainties, above it: the moment is twice the orbital value.
Answer. |μz| = (1.01 ± 0.04) μB and gₛ = 2.02 ± 0.08, consistent with 2.0023. A g = 1 moment would have split the deposits by only 0.074 mm, ruled out by about 13 standard uncertainties.
HardExample 3
Test the spinning-ball picture. Model the electron as a rigid sphere of radius r with its mass mₑ and charge −e both spread uniformly through it, spinning with angular momentum ħ/2. (a) Find the equator speed at the classical electron radius, 2.82 × 10⁻¹⁵ m. (b) Find the smallest radius at which that speed does not exceed c. (c) Integrate over the sphere to find its magnetic moment and g factor.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- A uniform sphere has I = (2/5)mₑr², so ħ/2 = (2/5)mₑr²ω, and the equator speed is veq = ωr = 5ħ/(4mₑr).
- (a) veq = 5 × 1.0546 × 10⁻³⁴ J s ÷ (4 × 9.109 × 10⁻³¹ kg × 2.818 × 10⁻¹⁵ m) = 5.14 × 10¹⁰ m s⁻¹, which is 171c. Using the full |S| = 0.866ħ in place of ħ/2 would make it 297c.
- (b) Set veq = c: r = 5ħ/(4mₑc) = 1.25 × 3.862 × 10⁻¹³ m = 4.83 × 10⁻¹³ m, 171 classical radii, and even then the equator moves at c, where the nonrelativistic I and L used here no longer hold.
- (c) Slice the sphere into thin rings about the spin axis. A ring of charge dq at distance s from the axis, turning at ω, is a current dq ω/2π around an area πs², so dμ = ½ωs² dq; its angular momentum is dL = ωs² dm.
- Charge follows mass, so dq = (−e/mₑ) dm in every ring and dμ/dL = −e/(2mₑ) ring by ring. Integrating both over the sphere keeps that ratio: μ = −(e/2mₑ)L for any radius and any density profile, provided charge follows mass.
- With Lz = ħ/2, μz = −eħ/(4mₑ) = −μB/2 = −4.64 × 10⁻²⁴ J T⁻¹, which is g = 1. The measured moment is −1.00116 μB, twice as large.
Answer. (a) 5.14 × 10¹⁰ m s⁻¹, about 171c; (b) r ≥ 4.83 × 10⁻¹³ m; (c) μz = −μB/2 = −4.64 × 10⁻²⁴ J T⁻¹, so g = 1, half the measured moment. The picture fails on speed and on moment.
Quick check
Select an answer and explain it before checking.
Review & continue
Electron spin is an intrinsic angular momentum with quantum number s = ½: its component along whatever axis you measure is only ever +ħ/2 or −ħ/2, and it carries a magnetic moment μ = −gₛ(e/2mₑ)S with gₛ ≈ 2.0023.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.